Question

Solve the following systems of equations.
$x^2+y^2=41,y-x=1$

Answer 1

Given that $x^2+y^2=41$ and $y-x=1$

Now we have,

$(y-x{)}^2=x^2+y^2-2\times x\times y$

$(1{)}^2=41-2xy$

$2xy=40$

$xy=20$

Now,

$(y+x{)}^2=(y-x{)}^2+4xy$

$(y+x{)}^2=(1{)}^2+4\times 20$

$(y+x{)}^2=81$

$(y+x)=9,-9$

Now we have

$(y-x)=\mathrm{1.......}\left(a\right)$

$(y+x)=\mathrm{9........}\left(b\right)$

Add $\left(a\right)$ and $\left(b\right)$, we have

$2y=10$

$y=5$

we have $y-x=1$

put the value of y on above equation

$x=4$

now $y+x=-9$ and$y-x=1$

add both the equation we have

$2y=-8$

$y=-4$

now we have $y-x=1$

put the value of y on above we have

$x=-5$ hence this the required solution